This page 10th cbse maths solution for exercise 4.3 part 4 is going to provide you solution for every problems that you find in the exercise no 4.3 part 4

(9) Two water taps together can fill a tank is 9** ⅜**
hours. The tap of the larger diameter takes 10 hours less than the
smaller one to fill the tank separately. Find the time in which each tap
can separately fill the tank.

Solution:

Let 'x' be the time taken by the smaller pipe to fill the part of tank

let 'x -10' be the time taken by the larger pipe to fill the part of the tank

Total time taken by two pipes = 9** ⅜**

part of tank filled by the smaller pipe in 1 hour = 1/x

part of tank filled by the larger pipe in 1 hour = 1/(x-10)

[1/x] + [1/(x - 10)] = 1/(9** ⅜)**

[1/x] + [1/(x - 10)] = 8/75

[(x - 10 + x)]/[x (x -10)] = 8/75

(2 x - 10)/(x**² - 10 x) = 8/75**

**75**(2 x - 10) = 8(x**² - 10 x)**

**150 x - 750 = 8 x****² - 80 x**

**8 x****² - 80 x**** - 150 x + 750 = 0**

**8 x****² - 230 x**** + 750 = 0**

**8 x****² - 200 x + 30 x**** + 750 = 0**

**8 x**** (x - 25) + 30 (x**** - 25) = 0**

**(8x + 30) (x -25) = 0**

** 8x + 30 = 0 x - 25 = 0**

** 8 x = -30 x = 25**

** x = -30/8**

** x = -15/4**

**Negative value is not acceptable **

**So time taken by smaller pipe alone to fill the tank = 25 hours**

**Time taken by larger pipe alone to fill the tank = 25 - 10 = 15 hours**

(10) A express train takes 1 hour less than a passenger train to travel 132 km between Mysore and Bangalore (without taking into consideration the time they stop at intermediate stations). If the average speed of the express train is 11 km/hr more than that of the passenger train, find the average speed of the two trains.

Solution:

Let 'x' be the average speed of passenger train

let 'x + 11' be the average speed of express train

Distance to be covered = 132 km

Time = Distance/speed

T₁ = 132/x

T₂= 132/(x + 11)

[132/x] - [132/(x + 11)] = 1

132[(1/x) - (1/(x + 11))] = 1

132 [ (x + 11 - x)/(x(x + 11))] = 1

132[11/(x**² + 11 x)] = 1**

** 1452/**(x**² + 11 x) = 1**

** 1452 = **x**² + 11 x**

x**² + 11 x - 1452 = 0**

x**² + 44 x - 33 x -1452 = 0**

x**(x + 44) - 33 (x + 44) = 0**

**(x + 44) (x - 33) = 0**

** x = -44 x = 33 **

**Speed should not be negative **

**So speed of passenger train = 33 km/hr**

**speed of express train = x + 11 = 33 + 11 = 44 km/hr**

(11) Sum of area of two squares is 468 m². If the difference of their perimeter is 24 m,find the sides of the two squares.

Solution:

Let 'x' be the side length of one square

let 'y' be the side length of another square

Sum of area of two squares = 468

x**² + y****² = 468 ------- (1)**

** 4 x - 4 y = 24**

** 4x = 24 + 4y **

** x = (24 + 4y)/4**

** x = (6 + y)**

Substitute x = 6 + y in the first equation

**(6 + y)****² + y****² = 468**

** 6****² + 2 (6)y + y****² + y****² = 468**

** 36 + 12 y + 2 y****² = 468**

** 2 y****² + 12 y - 468 + 36 = 0**

** ****2 y****² + 12 y - 432 = 0**

Dividing the whole equation by 2 =>

y² + 6 y - 216 = 0

y² + 18 y - 12 y - 216 = 0

y(y + 18) - 12(y + 18) = 0

(y - 12) (y + 18) = 0

y - 12 = 0 y + 18 = 0

y = 12 cm y = - 18

x = 6 + 12

x = 18 cm

Therefore the side length of two squares = 18 cm and 12 cm

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